Problem: $\dfrac{ 4x - 8y }{ 6 } = \dfrac{ -6x - 4z }{ -4 }$ Solve for $x$.
Answer: Multiply both sides by the left denominator. $\dfrac{ 4x - 8y }{ {6} } = \dfrac{ -6x - 4z }{ -4 }$ ${6} \cdot \dfrac{ 4x - 8y }{ {6} } = {6} \cdot \dfrac{ -6x - 4z }{ -4 }$ $4x - 8y = {6} \cdot \dfrac { -6x - 4z }{ -4 }$ Multiply both sides by the right denominator. $4x - 8y = 6 \cdot \dfrac{ -6x - 4z }{ -{4} }$ $-{4} \cdot \left( 4x - 8y \right) = -{4} \cdot 6 \cdot \dfrac{ -6x - 4z }{ -{4} }$ $-{4} \cdot \left( 4x - 8y \right) = 6 \cdot \left( -6x - 4z \right)$ Distribute both sides $-{4} \cdot \left( 4x - 8y \right) = {6} \cdot \left( -6x - 4z \right)$ $-{16}x + {32}y = -{36}x - {24}z$ Combine $x$ terms on the left. $-{16x} + 32y = -{36x} - 24z$ ${20x} + 32y = -24z$ Move the $y$ term to the right. $20x + {32y} = -24z$ $20x = -24z - {32y}$ Isolate $x$ by dividing both sides by its coefficient. ${20}x = -24z - 32y$ $x = \dfrac{ -24z - 32y }{ {20} }$ All of these terms are divisible by $4$ $x = \dfrac{ -{6}z - {8}y }{ {5} }$